Correct Answer - Option 4 : 1/9
Concept:
If an event E can occur in n(E)= k ways out of n(s)=n equally likely ways, then, \(\rm P \left( E \right) = \frac{{n\left( E \right)}}{{n\left( s \right)}} = \frac{k}{n}\).
Calculation:
The set, \(S = \left\{ \begin{array}{l}\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right)\\\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)\\\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right)\\\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)\\\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\)is a sample space for the given experiment.
There are 36 elements in the sample space, that is n(S) = 36.
If E is the event “sum is 5” then E = {(4,1), (3,2), (2,3), (1,4)}.
\(\rm P \left( {{\rm\text{sum is 5}}} \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{4}{{36}} = \frac{1}{9}\)
